A solid homogeneous sphere rolls without slip | vedclass

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Question

(A) For a body rolling without slipping down an inclined plane of height $h$,the conservation of mechanical energy gives: $mgh = \frac{1}{2}mv^2 + \fr

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$\sqrt{\frac{10}{7}gh}$

Vedclass · Answer

(A) For a body rolling without slipping down an inclined plane of height $h$,the conservation of mechanical energy gives: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since the body is a solid sphere,its moment of inertia about the center of mass is $I = \frac{2}{5}mR^2$. For rolling without slipping,$\omega = \frac{v}{R}$. Substituting these into the energy equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2$. $mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$. Solving for $v$,we get $v^2 = \frac{10}{7}gh$,which implies $v = \sqrt{\frac{10}{7}gh}$.

$A$ solid homogeneous sphere rolls without slipping down an inclined plane of height $h$. The velocity of the sphere at the bottom is:

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